The hybridisation of Ni in Ni-CO-4 is-sp-2dsp-2sp-3sp-3d

Question

Toppr Admin · Accepted Answer

The oxidation state of Ni is 0- The valence shell electronic configuration of the ground state of Ni is 3d84s2- but all the 10 electrons are pushed into-xA0-3d orbital due to the strong field CO ligands approaching the Ni atom- thereby forming sp3 hybridization-

The hybridisation of $N i$ in $N i (C O)_{4}$ is:
$s p^{2}$
$d s p^{2}$
$s p^{3}$
$s p^{3} d$

The oxidation state of $N i$ is 0. The valence shell electronic configuration of the ground state of $N i$ is $3 d^{8} 4 s^{2}$ , but all the 10 electrons are pushed into $3 d$ orbital due to the strong field $C O$ ligands approaching the $N i$ atom, thereby forming $s p^{3}$ hybridization.

The hybridisation of Ni in Ni(CO)4 is:sp2dsp2sp3sp3d

The oxidation state of Ni is 0. The valence shell electronic configuration of the ground state of Ni is 3d84s2, but all the 10 electrons are pushed into 3d orbital due to the strong field CO ligands approaching the Ni atom, thereby forming sp3 hybridization.

The hybridisation of $N i$ in $N i (C O)_{4}$ is:
$s p^{2}$
$d s p^{2}$
$s p^{3}$
$s p^{3} d$

The oxidation state of $N i$ is 0. The valence shell electronic configuration of the ground state of $N i$ is $3 d^{8} 4 s^{2}$ , but all the 10 electrons are pushed into $3 d$ orbital due to the strong field $C O$ ligands approaching the $N i$ atom, thereby forming $s p^{3}$ hybridization.